I am using strcmp in following ways
- Passing char[] array names
Passing pointers to string literals but, the second result in seg fault. even 开发者_如何转开发though i have confirmed that pointers point to correct string literals, i am confused as to why i am getting seg fault.. Here is the code:-
#include <stdio.h> #include <stdlib.h> #include <string.h> int main(int argc, char const *args[]) { char firstName[strlen(*++args)]; strcpy(firstName, *args); char lastName[strlen(*++args)]; strcpy(lastName, *args); printf("%s\t%s\n", firstName, lastName); printf("%d\n", strcmp(firstName, lastName));// this works printf("%d\n", strcmp(*(--args),*(++args)));//this gives me a seg fault return EXIT_SUCCESS; }
I am saving it as str.c and when i compile it, first i get following warning:
[Neutron@Discovery examples]$ gcc -Wall str.c -o str
str.c: In function ‘main’:
str.c:15: warning: operation on ‘args’ may be undefined
finally running it, gives a seg fault as shown below
[Neutron@Discovery examples]$ ./str Jimmy Neutron
Jimmy Neutron
-4
Segmentation fault (core dumped)
Don't use --
and ++
when you pass the same variable to the same function twice as two different parameters.
Instead of printf("%d\n", strcmp(*(--args),*(++args)));
do
char *first = *(--args);
char *second = *(++args);
printf("%d\n", strcmp(first,second));
Still not really readable (better use indexes and check against argc
for validity), but at least you don't change the value and evaluate multiple times it at the same sequence point.
In addition to what littleadv's post says, your buffers are one character too short (it didn't leave any space for the null-terminator). Thus, your strcpy
causes a buffer overflow.
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